EXERCISE 20 D
Question 1
Ans: Ans-1. We know that distance between two points
$\begin{aligned}&P\left(x_{1}, y_{1}\right) \text { and } Q\left(x_{2}, y_{2}\right) \\&=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}\end{aligned}$
(i) Distance between $(0,0)$ and $(2,3)$
$\begin{aligned}&=\sqrt{(2-0)^{2}+(3-0)^{2}}=\sqrt{2^{2}+3^{2}} \\&=\sqrt{4+9}=\sqrt{13}\end{aligned}$
(iii) istance between $(-3,0)$ and $(0, \sqrt{7})$
$\begin{aligned}&=\sqrt{[0-(-3)]^{2}+(\sqrt{7}-0)^{2}}=\sqrt{\left.(3)^{2}+\sqrt{7}\right)^{2}} \\&=\sqrt{9+7}=\sqrt{16}=4\end{aligned}$
(iv) Distance between $(7,9)$ and $(4,7)$
$\begin{aligned}&=\sqrt{(4-7)^{2}+(5-9)^{2}}=\sqrt{(-3)^{2}+(-4)^{2}} \\&=\sqrt{9+16}=\sqrt{25}=5\end{aligned}$
(v) Distance between $(-6,-1),(-6,11)$
$\begin{aligned}&=\sqrt{(-6-(-6)]^{2}+[11-(-1)]^{2}} \\&=\sqrt{(-6+6)^{2}+(11+1)^{2}}=\sqrt{0^{2}+12^{2}} \\&=\sqrt{12^{2}}=12\end{aligned}$
(vi)Distarce between $(a+b, a-b)$ and $(a-b,-a-b)$
$\begin{aligned}&=\sqrt{[(a-b)-(a+b)]^{2}+[(-a-b)-(a-b)]^{2}} \\&=\sqrt{(a-b-a-b)^{2}+(-a-b-a+b)^{2}} \\&=\sqrt{(-2 b)^{2}+(-2 a)^{2}} \\&=\sqrt{4 b^{2}+4 a^{2}}=\sqrt{4\left(a^{2}+b^{2}\right)} \\&=2 \sqrt{a^{2}+b^{2}}\end{aligned}$
(vii) Distance between $(2,-11)$ and $(-4,-3)$
$\begin{aligned}&=\sqrt{(-4-2)^{2}+(-3-(-11)]^{2}} \\&=\sqrt{(-6)^{2}+(-3+11)^{2}}=\sqrt{(-6)^{2}+\left(81^{2}\right.} \\&=\sqrt{36+64}=\sqrt{100}=10\end{aligned}$
(viii)
Distance between $(a, b)$ and (2a,b)
$\begin{aligned}&=\sqrt{(2 a-a)^{2}+(b-b)^{2}}=\sqrt{(a)^{2}+(0)^{2}} \\&=\sqrt{a^{2}}=a\end{aligned}$
Question 2
Ans-2. We know that distance between two points
$=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$
Distance between two points $(0,0)$ and $(x, 3)$
$\begin{aligned}&=\sqrt{(x-0)^{2}+(3-0)^{2}}=\sqrt{x^{2}+3^{2}} \\&=\sqrt{x^{2}+9}\end{aligned}$
But distance given $=5$
So $\sqrt{x^{2}+9}=5$
Question 3
Ans: (ii) Radius $=$ distance between $(0,01$ and $(-6,8)$
$\begin{aligned}&=\sqrt{(-6-0)^{2}+(8-0)^{2}}=\sqrt{(-6)^{2}+(8)^{2}} \\&=\sqrt{36+64}=\sqrt{100}=10 \text { units }\end{aligned}$
(ii)
Radius =distance $(2,0)$ and $(7,-12)$
$=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$
$=\sqrt{(7-2)^{2}+(-12-0)^{2}}$
$=\sqrt{(5)^{2}+(-12)^{2}}$
$=\sqrt{25+144}=\sqrt{169}=13$ units
Question 4
Ans: In a $\triangle A B C$ vertices are
$$\begin{aligned}&A(3,4), B(2,-1), C(4,-6) \\&\text { length of } A B=\sqrt{(2-3)^{2}+(-1-4)^{2}} \\&\left(Q \text {distance }=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}\right) \\&=\sqrt{(-1)^{2}+(-5)^{2}}=\sqrt{1+25}=\sqrt{26} \\&\text { length of } B C=\sqrt{\left.(4-2)^{2}+[-6+-1)\right]^{2}} \\&=\sqrt{(2)^{2}+(-6+1)^{2}}\end{aligned}$
$=\sqrt{(2)^{2}+(-5)^{2}}$
$=\sqrt{1+25}=\sqrt{2.3}$
and $C A=\sqrt{(4 \cdot 3)^{2}+(-6-1)^{2}}$
$\sqrt{(1)^{2}+(-10)^{2}}$
$=\sqrt{1+100}=\sqrt{101}$
Question 5
Ans: let the point on $x$-area be $(x, 0)$ as it lies $\mathrm{cm}$ $x$-axis there fore its y- Co-ordinate s =0
So $\begin{aligned} d &=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}} \\ \Rightarrow 10 &=\sqrt{(-4-x)^{2}+(8-0)^{2}} \\ \Rightarrow 10 &=\sqrt{(-4-x)^{2}+\left(81^{2}\right.} \\ \Rightarrow 10 &=\sqrt{16+x^{2}+8 x+64} \\ \Rightarrow 10 &=\sqrt{x^{2}+8 x+80} \end{aligned}$
Squaring both sides
$100=x^{2}+3 x+80$
$\Rightarrow x^{2}+8 x+80-100=0$
$\Rightarrow x^{2}+8 x-20=0$
$\Rightarrow x^{2}+10 x-2 x-20=0$
$\Rightarrow x(x+10)-2(x+10)=0$
$\Rightarrow(x+10)(x-2)=0$
Either x + 10 = 0 Then x =-10
or $x-2=0$, then $x=2$
So the points will be (-10, 0) and (2,0)
Question 6
Ans: Let point A be the required point which is equidistant from P and Q
if it lies on y - axis
So it x - coordinate =0
Let point A be (o, y)
Using distance formula
Distance between AP and AQ are equal
So $\begin{aligned} & \sqrt{(0-0)^{2}+(y-8)^{2}}=\sqrt{\left[0-(-4)^{2}+(y-4)^{2}\right.} \\ \Rightarrow & \sqrt{0^{2}+(y-8)^{2}}=\sqrt{(4)^{2}+(y-4)^{2}} \\=& \sqrt{(y-8)^{2}}=\sqrt{16+(y-4)^{2}} \end{aligned}$
squaring boths sides
$\begin{aligned}&(y-8)^{2}=16+(y-4)^{2} \\&y^{2}-16 y+64=16+y^{2}-8 y-16 \\&y^{2}-16 y-y^{2}+8 y=32-64 \\&-8 y=-32 \\&\Rightarrow y=\frac{-32}{-8}=4\end{aligned}$
So points will be $(0,4)$
Question 7
Ans: Let length of line AD = 10
One end points $A(-3,2)$ and let second end points is be $(x, 10)$
$\text { So } \begin{aligned}A B &=\sqrt{[x-(-3)]^{2}+(10-2)^{2}} \\=10 &=\sqrt{(x+3)^{2}+(3)^{2}}\end{aligned}$
squaring both sides
$\begin{aligned}&100=(x+3)^{2}+64 \\&(x+3)^{2}+64-100=0 \\&x^{2}+6 x+9+64-100=0 \\&x^{2}+6 x-27=0 \\&\Rightarrow x^{2}+9 x-3 x-27=0 \\&\Rightarrow x(x+9)-3(x+9)=0 \\&\Rightarrow(x+9)(x-3)=0\end{aligned}$
Either $x+9=0$ then $x=-9$
or $x-3=0$ then $x=3$
Abscissa $=-9$ or 3 Hence proved
Question 8
Ans: (i) let $A(-5,1) \quad B(1,-1)$ and $C(1,-2)$ are the points
If sum of length of any two lines is equal to the third then these points are collinear
Now AB= $\sqrt{(1+5)^{2}+(-1-1)^{2}}$
$=\sqrt{6^{2}+(-2)^{2}}=(\sqrt{36+4}=\sqrt{40}=2 \sqrt{10}$
$B C=\sqrt{(1-1)^{2}+(-2+1)^{2}}$
$=\sqrt{0^{2}+(-1)^{2}}=\sqrt{(-1)^{2}}=\sqrt{1}=1$
$C A=\sqrt{(1+5)^{2}+(-2-1)^{2}}=\sqrt{6^{2}+(-3)^{2}}$
$=\sqrt{36+9}=\sqrt{45}$
$=\sqrt{9 \times 5}=3 \sqrt{5}$
We see that the points are not collinear
(ii) if points $(-1,3)$, $(2, P)$ ard $(5,-1)$ are collinear
Let $A(-1,3)$ B (2,P ) andC (5,-1)
Now $A A=\sqrt{(-1-2)^{2}+(3-p)^{2}}=\sqrt{(-3)^{2}+(3-p)^{2}}$
$=\sqrt{9+9-6 p+p^{2}}=\sqrt{p^{2}-6 p+18}$
$B C=\sqrt{(2-5)^{2}+(p+1)^{2}}$
$=\sqrt{(-3)^{2}+(p+1)^{2}}$
$=\sqrt{9+p^{2}+2 p+1}$
$=\sqrt{p^{2}+2 p+10}$
and $C A=\sqrt{(5+1)^{2}+(-1-3)^{2}}$
$=\sqrt{(6)^{2}+(-4)^{2}}=\sqrt{36+16}=\sqrt{52}$
if $A , B$ and $C$ orre collincar
so $B C=A B+C A$
$\sqrt{p^{2}+2 p}+10=\sqrt{p^{2}-6 p+18}+\sqrt{52}$
squaring both sides
$\begin{aligned}&p^{2}+2 p+10=p^{2}-6 p+18+52+2 \sqrt{52} \\&\sqrt{p^{2}-6 p+18}\end{aligned}$
$p^{2}+2 p+10-p^{2}+6 p-18-52=2 \times 2 \sqrt{13}$
$\sqrt{p^{2}-6 p+18}$
$8 p-60=4 \sqrt{13} \sqrt{p^{2}-6 p+18}$
$2 p-15=\sqrt{13\left(p^{2}-6 p+18\right.}$
Again squaring both sides
$(2 p-15)^{2}=13\left(p^{2}-6 p+18\right)$
$\Rightarrow 4 p^{2}-60 p+225=13 p^{2}-78 p+234$
$\Rightarrow 13 p^{2}-78 p+234-4 p^{2}+60 p-225=0$
$\Rightarrow 9 p^{2}-18 p+9=0$
$\Rightarrow p^{2}-2 p+1=0$
$\Rightarrow(p-1)^{2}=0 \Rightarrow p-1=0$
so $p=1$
Question 9
Ans: Points are $(-5,2), B(1,-2)$ and $C(x, 4)$
$\begin{aligned}&\text { Now } A B=\sqrt{(-5-1)^{2}+(2+2)^{2}} \\&=\sqrt{(-6)^{2}+(4)^{2}}=\sqrt{36+16}=\sqrt{52} \\&B C=\sqrt{(1-x)^{2}}+\left(-2-4^{2}\right.) \\&=\sqrt{(1-x)^{2}+(6)^{2}}=\sqrt{(1-x)^{2}+36}\end{aligned}$
if $A B=B C$
so $\sqrt{52}=\sqrt{(1-x)^{2}+36}$
Squaring both sides
$\begin{aligned}&52=(1-x)^{2}+36 \\&\Rightarrow(1-x)^{2}=52-36=16\end{aligned}$
$\Rightarrow 1+x^{2}+2 x=16 \Rightarrow x^{2}+2 x+1-16=0$
$\Rightarrow x^{2}+2 x-15=0$
$\Rightarrow x^{2}+5 x-3 x-15=0$
$\Rightarrow x(x+5)-3(x+5)=0$
$\Rightarrow(x+5)(x-3)=0$
Either $x+5=0$ then $x=-5$
or $x-3=0$, then $x=3$
Hence $x=3,-5$
Question 10
Ans: We know that a triangle
Is an isosceles if any two sides are equal ,
Now $P(2,0), Q(6,0)$ and $R(4,4)$
Now $\begin{aligned} P Q &=\sqrt{(6-2)^{2}+(0-0)^{2}}=\sqrt{4^{2}+0^{2}} \\ &=\sqrt{16+0}=\sqrt{16}=4 \end{aligned}$
$Q R=\sqrt{(6-412}+(0-4)^{2}=\sqrt{(2)^{2}+(-4)^{2}}$
$=\sqrt{4+16}=\sqrt{20}$
$R P=\sqrt{(4-2)^{2}+(4-0)^{2}}=\sqrt{(2)^{2}+(4)^{2}}$
$=\sqrt{4+16}=\sqrt{20}$
if $Q R=R P=\sqrt{20}$
So $\triangle P Q R$ is an isosceles triangle.
Question 12
Ans:(i) points are $A(0,4) B(2,5) C(3,3)$
$\begin{aligned}A B &=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}} \\&=\sqrt{(2-0)^{2}+(5-4)^{2}}=\sqrt{(2)^{2}+(1)^{2}} \\&=\sqrt{4+1}=\sqrt{5}\end{aligned}$
Similarly $B C=\sqrt{(3-2)^{2}+(3-5)^{2}}$
$=\sqrt{(1)^{2}+(-2)^{2}}+\sqrt{1+4}=\sqrt{5}$
and $\begin{aligned} C A &=\sqrt{(0-3)^{2}}+\left(4-31^{2}\right.\\=& \sqrt{(-3)^{2}+(1)^{2}}=\sqrt{9+1}=\sqrt{10} \end{aligned}$
We see that AB = BC = $\sqrt{5}$
So it is an isosceles triangle
And $A B^{2}+B C^{2}=(\sqrt{5})^{2}+(\sqrt{5})^{2}=5+5=10$
and $C A^{2}=\left(\left.\sqrt{10}\right)^{2}=10\right.$
if $A B^{2}+B C^{2}=C A^{2}$
So it is an isosceles right angle
Area of $\triangle A B C=\frac{1}{2} A B \times B C$
$=\frac{1}{2} \sqrt{5} \times \sqrt{5}=\frac{5}{2}=2.5$ square units
(ii)points are $A(3,5), B(1,1), C(5,3)$ and $D(7,7)$
$\begin{aligned}A B &=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}} \\&=\sqrt{(1-3)^{2}+(1-5)^{2}}=\sqrt{(-2)^{2}+(-4)^{2}} \\&=\sqrt{4+16}=\sqrt{20}\end{aligned}$
$\begin{aligned} B C &=\sqrt{(3+)^{2}+(3-1)^{2}}=\sqrt{(4)^{2}+(2)^{2}} \\ &=\sqrt{16+4}=\sqrt{20} \\ C D &=\sqrt{(7-5)^{2}+(7-3)^{2}}=\sqrt{\left(2.1^{2}+(4)^{1}\right.} \\ &=\sqrt{4+16}=\sqrt{2 0} \end{aligned}$
$D A=\sqrt{(3-7)^{2}+(5-7)^{2}}$
$=\sqrt{(-4)^{2}+(-2)^{2}}=\sqrt{16+4}=\sqrt{20}$
we see that $A B=B C=C D=D A$
So ABCD is a rhombus or a square
Not diagonal AC = $\sqrt{(5-3)^{2}+(3-5)^{2}}$
$=\sqrt{(2)^{2}+(-2)^{2}}=\sqrt{4+4}=\sqrt{8}$
and dingonal BD $=\sqrt{(7-1)^{2}+(7-1)^{2}}$
$=\sqrt{(6)^{2}+(6)^{2}}=\sqrt{36+36}=\sqrt{72}$
If Diagonals are not equal
So it is a rhombus not a square
Not area of rhombus ABCD= $\frac{1}{2}$ product of Diagonals
$=\frac{1}{2} \sqrt{8} \times \sqrt{72}$
$=\frac{1}{2} \sqrt{576}$
$=\frac{1}{2} \times 24=12 \mathrm{sq} .$ units
Question 13
Ans: (i) Points are $A(1,5), B(3,5)$ C (2,$\sqrt{3}$ )
Now $A B=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$
$=\sqrt{(3-1)^{2}+(\sqrt{3}-\sqrt{3})^{2}}=\sqrt{(2)^{2}+(0)^{2}}$
$=\sqrt{4+0}=\sqrt{4}=2$
Similarly BC = $\sqrt{(2-3)^{2}+}$ 2$\sqrt{3}$ - $\sqrt{3}^{2}$
$=\sqrt{(1)^{2}+(13)^{2}}=\sqrt{1+3}=\sqrt{4}=2$
and $C A=\sqrt{(1-2)^{2}}+(\sqrt{3}-2 \sqrt{3})^{2}$
$=\sqrt{(-1)^{2}+(\sqrt{-3})^{2}}$
$=\sqrt{1+3}=\sqrt{4}=2$
We see that AB = BC = CA=2
So $\triangle A B C$ is an equilateral triangle
(ii) points $A(1.1) \quad B(-1,-1) \cdot C$ (-$\sqrt{3}$ ,$\sqrt{3}$)
Now $\begin{aligned} A D &=\sqrt{\left(x_{2}-x_{1}\right)^{2}}+\left(y_{2}-y_{1}\right)^{2} \\=& \sqrt{(-1-1)^{2}+(-1-1)^{2}} \end{aligned}$
$=\sqrt{(-2)^{2}+(-2)^{2}}=\sqrt{4+4}=\sqrt{8}$
similarly $B C=\sqrt{[-53-(-1)]^{2}+\left[\sqrt{3}-(-1]^{2}\right.}$
$=\sqrt{(-\sqrt{3}+1)^{2}+(\sqrt{3}+1)^{2}}$
$\begin{aligned}=& \sqrt{(3+1-25+3+1+2 \sqrt{3})} \\ &=\sqrt{8} \end{aligned}$
We see that AB = BC = CA = $\sqrt{8}$
So $\triangle A B C$ is an equilateral triangle.
Question 14
Ans: Let the vertices be $A(-2,61)(5,3), C(-1,-11)$ and $D(-8,-8)$
$\begin{aligned} A B=& \sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}} \\=& \sqrt{(5-(-2))^{2}+(3-6)^{2}}=\sqrt{(5+2)^{2}+(3-6)^{2}} \\=& \sqrt{(7)^{2}+(-3)^{2}}=\sqrt{49+9}=\sqrt{58} \\ & \text { similanly } B C=\sqrt{(-1-5)^{2}+(-11-3)^{2}} \end{aligned}$
$=\sqrt{(-6)^{2}+(-14)^{2}}$
$=\sqrt{36+196}=\sqrt{232}$
$C D=\sqrt{\left[-8-(-11]^{2}+[-8-(-11)]^{2}\right.}$
$=\sqrt{(-8+1)^{2}+(-8+11)^{2}}$
$=\sqrt{(-7)^{2}+(3)^{2}}=\sqrt{49+19}=\sqrt{58}$
$\begin{aligned} O A &=\sqrt{-2-(-8)^{2}+[6-(-8)]^{2}} \\ &=\sqrt{(-2+8)^{2}+(6+8)^{2}} \\ &=\sqrt{(6)^{2}+(14)^{2}}=\sqrt{36+196}=\sqrt{232} \end{aligned}$
We see that AB = CD and AD = BC
So ABCD is a rectangle
Question 15
Ans: Center of the circle is origin $0(0,0)$ and length of the radius of the circle $=10$ units
(i) Now distance between $(0,0)$ and $(6,81$
$\begin{aligned}&=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}} \\&=\sqrt{(6-0)^{2}+(8-0)^{2}}=\sqrt{(6)^{2}+(8)^{2}} \\&=\sqrt{36+64}=\sqrt{100}=10\end{aligned}$
So This point $(6,8)$ lies on the circle
(ii) Similarly distance between $(0,0)$ and $(0,11)$
$\begin{aligned}&=\sqrt{(0-0)^{2}+(11-0)^{2}}=\sqrt{(0)^{2}+(11)^{2}} \\&=\sqrt{0+121}=\sqrt{121}=11 \\&\text { is } 11>10\end{aligned}$
so This point is outside the circle
(iii) Distance between $(0,0)$ and $(-10,10)$
$\begin{aligned}&=\sqrt{(-10-0)^{2}+(0-0)^{2}} \\&=\sqrt{(-10)^{2}+101^{2}}=\sqrt{100+0}=\sqrt{100}=10\end{aligned}$
so This point lies on the circle
(iv) Distance between $(0,0)$ and $(7,7)$
$\begin{aligned}&=\sqrt{(7-0)^{2}+\left(7-01^{2}\right.}=\sqrt{(7)^{2}+(7)^{2}} \\&=\sqrt{49+49}=\sqrt{98}\end{aligned}$
So $\sqrt{98}<10$
So This point is inside the circle
(v) Distance between $(0.0)$ and $(-9,+1$
$\begin{aligned}&=\sqrt{(-9-0)^{2}+(4-0)^{2}} \\&=\sqrt{(-9)^{2}+(4)^{2}} \\&=\sqrt{81+16}=\sqrt{97} \\&\quad \text { if } \sqrt{97}<10\end{aligned}$
So This point is inside the circle
Question 16
Ans: if the point $p(2,-5)$ is mapped into point p' in the $x$-axis
So The co- ordinates of p' will be (2,5)
If the point is (3,7) is mapped to 81 in the origin
so co-ordinates of $Q^{\prime}$ will be $(-3,-71$
Now length of $P Q=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$
$\begin{aligned}&=\sqrt{(3-2)^{2}+(7-(-5)]^{2}} \\&=\sqrt{(1)^{2}+(7+5)^{2}}=\sqrt{(1)^{2}+(12)^{2}} \\&=\sqrt{1+(44}=\sqrt{145 \text { units }} \\&\text { and } \left.p^{\prime} \theta\right)=\sqrt{(-3-2)^{2}+\left(-7,-3^{2}\right.}\end{aligned}$
$=\sqrt{(-5)^{2}+(-12)^{2}}$
$=\sqrt{25+144}=\sqrt{169}=13$ units
Question 17
Ans: Co-ordinates of points $P$ and $Q$ are $(4,1)$ and $(2,0)$ respectively.
(i) p' is the image of $p(4,1)$ under reflection in $y$.ary is as so $c o$-ordinates of p' will be $(-4,1)$
(ii) The line joining the point p and p' is parallel to to x- axis at a distance of 1 on the positive of side y-axis
If Q' is the image of $Q(2,0)$ in the line pp'
So $(2,2 x-y)$ or $2(2 \times 1-0)$ or $(2,2-01$ or $(2,2)$
(iii)
$\begin{aligned}\| v_{0}^{\prime} &=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}} \\&=\sqrt{\left[2-(-4)^{2}+(2-1)^{2}\right.} \\&=\sqrt{\left.(2+4)^{2}+\pi\right)^{2}}=\sqrt{(6)^{2}+\left(11^{2}\right.} \\&=\sqrt{36+1}=\sqrt{37} \text { units }\end{aligned}$
Question 18
Ans: $A(5,1)$ in the center of circle Radius as the circle of $=13$ units $A B \perp P Q$ where $P Q$ is a chord al the circle Co-ordinates of $D$ are $(2,-3)$
if $A B \perp P Q$
So $B$ is the mid-point of $P Q$
(i)
Now AD $\begin{aligned} &=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}} \\ &=\sqrt{(2-5)^{2}+(-3-1)^{2}} \\ &=\sqrt{\left.(-3)^{2}+1-4\right)^{2}}=\sqrt{9+16} \\ &=\sqrt{25}=5 \text { units } \end{aligned}$
(ii) $\triangle A P B .$
$A P^{2}=A B^{2}+P B^{2} \Rightarrow(13)^{2}=(5)^{2}+P B^{2}$
$=169=25+P B^{2} \Rightarrow P B^{2}=169-25=144$
$=(12)^{2}$
So $P B=12$ units
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